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Question: Solve the following questions: (i) Van’t Hoff-Boyle’s law (ii) Van’t Hoff-Charle’s...

Solve the following questions:
(i) Van’t Hoff-Boyle’s law (ii) Van’t Hoff-Charle’s

Explanation

Solution

Hint: You need to recall the laws from the kinetic theory of gases. Once you get the laws try to relate them with the ideal gas equation. Now you can easily explain these given laws to reach your answer.

Complete step by step answer:

We will discuss these laws one by one -

Van't Hoff Boyle's Law of the solution:
At constant temperature, the osmotic pressure (π\pi) of a dilute solution is directly proportional to its molar concentration (C) or inversely proportional to volume (V) of the solution.
Explanation:
From the equation, π\pi = CRT
As we know, π\pi \propto C but C = n/V
So we can write, π\pi \propto n/V
or π\pi \propto 1/V
Or we can say π\piV = constant
[Here, n is no. of moles of solute dissolved in liters ]

Van't Hoff Charle's Law of the solution: The concentration remaining constant, the osmotic pressure (π\pi) of a dilute solution is directly proportional to absolute temperature (T) of the solution.
Explanation:
From the equation, π\pi = CRT
As we know, π\pi \propto T
So, we can write,
π\pi/T = constant.

Therefore, we explained both of these laws.

Note: There is another law related to the laws we mentioned above -
Van't Hoff-Avogadro's law: At a given temperature, equal volumes of the solutions having the same osmotic pressure contain an equal number of solute particles.
Explanation:
One mole of a substance contains Avogadro's number of molecules. Therefore, the equal number of moles of the substances will contain an equal number of molecules.
The osmotic pressure, π\pi, is given by π\pi = CRT
where C is the concentration of a solution in mol dm3dm^{ -3 }.
π\pi = CRT
We can write π\pi \propto nV at a constant temperature. (C = n/V)
If two solutions contain nA and nB moles of two solutes in the equal volumes (V) of the solutions A and B at the same temperature, then πAnA/V{ \pi }_{ A }\quad \propto \quad { n }_{ A }/V and πBnB/V{ \pi }_{ B }\quad \propto \quad { n }_{ B }/V
Therefore, πAπB=nAnB{ \pi }_{ A }{ \pi }_{ B }\quad =\quad { n }_{ A }{ n }_{ B }
If πA=πB{ \pi }_{ A }{ \quad =\quad \pi }_{ B } then
nA=nBn_{ A }=\quad n_{ B }